# FULL BREAKDOWN OF TIER 4 RIFTS

Milan1405
Posts:

**75**
So I decided to make a guide on all of the probabilities of getting certain rewards. This took me a long time so I hope you find it useful. Enjoy

19

Milan1405
Posts: **75**

So I decided to make a guide on all of the probabilities of getting certain rewards. This took me a long time so I hope you find it useful. Enjoy

19

## Comments

303385179976816754962957547538956567,215Let's start by opening one crystal. The odds of getting the AG are 5%, which is another way of saying there's five out of one hundred chance to get the AG. So if you were to open a hundred crystals then on average you'd expect to get five AGs and 95 something else. Imagine a room full of players. All of them open one crystal. 5% of them get an AG, 95% don't (on average). Imagine all the winners leave that room and enter another room: the winners' room. So after opening one crystal 95% of the players are still in the original room. Now we open a second crystal. 5% of those players should get an AG (on average), and 95% should miss again. 5% of *those* players are now winners and leave the room. How many was that? Well, it was 5% of 95% of the original, which is 4.75% of the original. So the winners' room has 5% + 4.75% = 9.75% of the original players. Meanwhile, 95% of 95% are still in the original room, which is 90.25% of the original players.

See the pattern? First, there's 95% left (the "losers"). Then there's 95% x 95% = 90.25% left. Then there will be 95% x 95% x 95% = 85.74%. The "losers" are 95% x 95% x ... 95% N times after you open X crystals, on average. So if you open eight crystals, the number of players in the losers room will be 95% x 95% x 95% x 95% x 95% x 95% x 95% x 95% = 0.95 ^ 8 = 66.34% of the original players.

If 66.34% of the original players are still in the original (losers) room because they haven't pulled an AG, then that means 33.66% of the original players are in the winners' room because they did pull an AG at some point. The statistical formulas are mathematical short hand for doing this kind of counting.

92if you run T4 Unstable 3x, you have a 5% chance to get the 5* AG, 3 times.

if you run T4 Unstable 1000x, you have a 5% chance to get the 5* AG, 1000 times

92You have a 99% chance to lose.

The chance you lose twice in a row is 99% or 99%, or .99*.99

The chance you lose three times in a row is .99 *.99 *.99.

The chance you lose 100 times in a row is .99 times itself 100 times

That comes to .36 or so.

So, if your chance to lose 100 in a row is .36, your chance to win at some point in that run is .64 (found by taking 1-.36)

that's how he's getting a 40% chance to get a 5* AG from running T4 Unstable 10x.

389In this case X is 0.95 because is the chance of it not being the gem and Y is 8 tries. Sorry I know it is confusing but it is hard to try to explain this on my phone

389757,215Where you might use a Markov chain is to analyze a situation like this: I save 5* shards until the next featured crystal arrives at a rate of 20k per week. I will open featured crystals until I get a featured champ, then open basic crystals until I run out of shards, unless I awaken a god tier champ, in which case I will stop until the next featured crystal arrives. What's the average number of shards I'm likely to have at any moment in time over the long run? How often will I run out?

Google isn't modelled as a Markov chain, that's kind of a weird statement. I believe you're referring to the Pagemark algorithm which forms the original heart of the Google system. Pagemark models

peopleas random web browsers that read a page, randomly click on a link on that page to go to another page, and repeat until they get bored and restart the process on another random page. In other words, tvtropes and wikipedia hell. This can be analyzed as a stochastic Markov process to find steady state transition vectors that assign weights to every web page that can be used to measure how likely a person would be to find and read the page, and use that as a measure of "importance." Important pages are those that more people read, or more web authors link to. Pagemark maps the problem of finding a solution to the problem of assigning page weights to a Markovian analysis.See: http://blog.kleinproject.org/?p=280

3897521217547,215The basic idea is that it can be difficult to properly express all the conditions and transitions in equation or matrix form. But it can be easier to express them algorithmically. So you write a program (or script) that basically goes step by step and simulates opening crystals and burning shards over a period of simulated time, all the while adding crystals, adding pretend champions to the game, etc. That's just step by step lines of code. Then you run it a million times and average the results. That will give you an estimate for what the actual values are in steady state.

An example of the kind of analysis I'd use Monte Carlo Markov for is comparing the signature ability of Red Magneto and White Magneto. Red Magneto has a safeguard that is always on while White Magneto has a safeguard that only triggers when low health but is stronger than Red Magneto's. So which one is better? Well, they are different so there's no one correct answer, but you can ask questions to try to compare them, like "given certain reasonable assumptions about the strength of the attacker, which one is harder/takes longer to kill?" And that's something that tends to be easier to solve by Monte Carlo Markov analysis. Especially because if you make a mistake or want to redo the analysis with different assumptions, instead of doing a whole bunch of math again you just tweak some numbers and rerun your script, and grab a beer while you wait.

389